JEE MainPhysics

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Pitch of a Helical Path in a Uniform Magnetic Field

A proton with a kinetic energy of 2.0eV moves into a region of uniform magnetic field of magnitude π/2× 10^(-3)T. The angle between the direction of magnetic field and velocity of proton is 60°. The pitch of the helical path taken by the proton is ______cm. (Take, mass of proton = 1.6× 10^(-27)kgand Charge on proton = 1.6× 10^(-19)C).

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Step-by-step solutionView

Correct answer

The pitch of the helical path traversed by the proton is 40 cm.

Option analysis

Why each option works or fails

Reviewed route

Solution

StepWorking

01Given

$K = 2.0 eV = 2.0 \times 1.6 \times 10^{- 19} J$ , $B = \frac{π}{2} \times 10^{- 3} T$ , $θ = 60^{\circ}$ , $m = 1.6 \times 10^{- 27} kg$ , $q = 1.6 \times 10^{- 19} C$ .

02Find

Pitch of the helical path $p$ in centimeters.

03Visualise

The velocity vector $\vec{v}$ is inclined at $60^{\circ}$ to the magnetic field $\vec{B}$ . The perpendicular component $v_{⊥} = v \sin 60^{\circ}$ provides circular motion, while the parallel component $v_{∥} = v \cos 60^{\circ}$ translates the particle along $\vec{B}$ , tracing a helix.

04Strategise

The pitch is the distance traveled along the magnetic field during one full time period of revolution: $p = v_{∥} T = (v \cos θ) (\frac{2 π m}{q B})$ , where $v = \sqrt{\frac{2 K}{m}}$ .

05Execute

First, calculate velocity $v = \sqrt{\frac{2 \times 2.0 \times 1.6 \times 10^{- 19}}{1.6 \times 10^{- 27}}} = \sqrt{4 \times 10^{8}} = 2 \times 10^{4} m/s$ . Next, the time period is $T = \frac{2 π \times 1.6 \times 10^{- 27}}{1.6 \times 10^{- 19} \times (π / 2 \times 10^{- 3})} = \frac{2 π \times 10^{- 8}}{(π / 2) \times 10^{- 3}} = 4 \times 10^{- 5} s$ . Then, pitch $p = v \cos 60^{\circ} \times T = (2 \times 10^{4}) \times 0.5 \times (4 \times 10^{- 5}) = 0.4 m = 40 cm$ .

✓Verify

Units: $(m/s) \times s = m = 100 cm$ . At non-relativistic speed ( $20 km/s ≪ c$ ), the cyclotron formula holds accurately.

Hints that build this answer step by step

Which formula correctly gives the pitch $p$ of a helical trajectory of a charged particle in a uniform magnetic field?
$p = v_{∥} T = (v \cos θ) (\frac{2 π m}{q B})$
What is the speed $v$ of the proton with kinetic energy $K = 2.0 eV$ ?
$v = \sqrt{\frac{2 \times (2.0 \times 1.6 \times 10^{- 19} J)}{1.6 \times 10^{- 27} kg}} = 2.0 \times 10^{4} m/s$
Substituting $v = 2.0 \times 10^{4} m/s$ , $θ = 60^{\circ}$ , and $B = \frac{π}{2} \times 10^{- 3} T$ , what is the value of the pitch in centimeters?
$40 cm$

✓ Source and academic review

Question type: Numerical
Exam relevance: JEE Main · Physics
Academic status: Answer verified against the official NTA key
Source: Previous-year question
Editorial review: 19 September 2026

Quick checks

Students also ask

Why is pitch dependent on v*cos(θ) instead of v*sin(θ)?

Pitch is the linear distance advanced along the magnetic field line in one revolution. The velocity component along the field line is parallel, which is $v \cos θ$ . The perpendicular component $v \sin θ$ governs the radius of the circle, not the translation along the axis.

Answer

Accepted answer: 40 cm

The pitch of the helical path traversed by the proton is 40 cm.

Why each option works or fails

Step-by-step solution

Given: $K = 2.0 eV = 2.0 \times 1.6 \times 10^{- 19} J$ , $B = \frac{π}{2} \times 10^{- 3} T$ , $θ = 60^{\circ}$ , $m = 1.6 \times 10^{- 27} kg$ , $q = 1.6 \times 10^{- 19} C$ .
Find: Pitch of the helical path $p$ in centimeters.
Visualise: The velocity vector $\vec{v}$ is inclined at $60^{\circ}$ to the magnetic field $\vec{B}$ . The perpendicular component $v_{⊥} = v \sin 60^{\circ}$ provides circular motion, while the parallel component $v_{∥} = v \cos 60^{\circ}$ translates the particle along $\vec{B}$ , tracing a helix.
Strategise: The pitch is the distance traveled along the magnetic field during one full time period of revolution: $p = v_{∥} T = (v \cos θ) (\frac{2 π m}{q B})$ , where $v = \sqrt{\frac{2 K}{m}}$ .
Execute: First, calculate velocity $v = \sqrt{\frac{2 \times 2.0 \times 1.6 \times 10^{- 19}}{1.6 \times 10^{- 27}}} = \sqrt{4 \times 10^{8}} = 2 \times 10^{4} m/s$ . Next, the time period is $T = \frac{2 π \times 1.6 \times 10^{- 27}}{1.6 \times 10^{- 19} \times (π / 2 \times 10^{- 3})} = \frac{2 π \times 10^{- 8}}{(π / 2) \times 10^{- 3}} = 4 \times 10^{- 5} s$ . Then, pitch $p = v \cos 60^{\circ} \times T = (2 \times 10^{4}) \times 0.5 \times (4 \times 10^{- 5}) = 0.4 m = 40 cm$ .
Verify: Units: $(m/s) \times s = m = 100 cm$ . At non-relativistic speed ( $20 km/s ≪ c$ ), the cyclotron formula holds accurately.

Pitch of a Helical Path in a Uniform Magnetic Field

What feels right?

The pitch of the helical path traversed by the proton is 40 cm.

Why each option works or fails

Solution

01Given

02Find

03Visualise

04Strategise

05Execute

✓Verify

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01Given

02Find

03Visualise

04Strategise

05Execute

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